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Ch. 5 - Systems and Matrices
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
All textbooksLial 13th EditionCh. 5 - Systems and MatricesProblem 23
Chapter 6, Problem 23

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (-3)/(x2(x2 + 5))

Verified step by step guidance
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Identify the denominator and factor it completely. The denominator is \(x^2(x^2 + 5)\), which consists of a repeated linear factor \(x^2\) and an irreducible quadratic factor \(x^2 + 5\).
Set up the form of the partial fraction decomposition. For the repeated linear factor \(x^2\), include terms with powers 1 and 2 in the denominator: \(\frac{A}{x} + \frac{B}{x^2}\). For the irreducible quadratic \(x^2 + 5\), include a linear numerator: \(\frac{Cx + D}{x^2 + 5}\).
Write the equation expressing the original fraction as the sum of the partial fractions: \(\frac{-3}{x^2(x^2 + 5)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 5}\).
Multiply both sides of the equation by the common denominator \(x^2(x^2 + 5)\) to clear the denominators, resulting in a polynomial equation: \(-3 = A x (x^2 + 5) + B (x^2 + 5) + (Cx + D) x^2\).
Expand the right side, collect like terms by powers of \(x\), and then equate the coefficients of corresponding powers of \(x\) on both sides to form a system of equations. Solve this system to find the values of \(A\), \(B\), \(C\), and \(D\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a complex rational expression as a sum of simpler fractions. This technique is especially useful for integrating rational functions or solving algebraic equations. The goal is to break down the given fraction into components with simpler denominators.
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Decomposition of Functions

Factoring the Denominator

To perform partial fraction decomposition, the denominator must be factored into irreducible polynomials. In this problem, the denominator is x²(x² + 5), which includes a repeated linear factor (x²) and an irreducible quadratic factor (x² + 5). Recognizing these factors guides the form of the decomposition.
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Rationalizing Denominators

Form of Partial Fractions for Repeated and Quadratic Factors

For repeated linear factors like x², the decomposition includes terms with denominators x and x². For irreducible quadratic factors like x² + 5, the numerator is a linear expression (Ax + B). Understanding these forms helps set up the correct equation to solve for unknown coefficients.
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Related Practice
Textbook Question

Solve each system by elimination. In systems with fractions, first clear denominators.

2x - 3y = -7

5x + 4y = 17

Textbook Question

Evaluate each determinant.

∣120−12−1014∣\(\left\)| \(\begin{matrix}\) 1 & 2 & 0 \\ -1 & 2 & -1 \\ 0 & 1 & 4 \(\end{matrix}\) \(\right\)|

Textbook Question

Find the inverse, if it exists, for each matrix.[1010−10211]\(\left\)[ \(\begin{matrix}\) 1 & 0 & 1 \\ 0 & -1 & 0 \\ 2 & 1 & 1 \(\end{matrix}\) \(\right\)]

Textbook Question

Solve each system by elimination. In systems with fractions, first clear denominators.

5x + 7y = 6

10x - 3y = 46

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Textbook Question

Solve each nonlinear system of equations. Give all solutions, including those with nonreal complex components. See Examples 1–5.

x2+y2=5x^2+y^2=5

−3x+4y=2-3x+4y=2

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Textbook Question

Use the Gauss-Jordan method to solve each system of equations. For systems in two variables with infinitely many solutions, write the solution with y arbitrary. For systems in three variables with infinitely many solutions, write the solution set with z arbitrary.

3x + 2y = -9

2x - 5y = -6

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